Find normality of 3 M of (al2so4)3
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ExpTotal no of moles of ions in solution is 0.1 M of CuSO
4
which gives mole of Cu
2+
=0.1, mole of SO
4
2−
=0.1
Also, 0.1 M of Al
2
(SO
4
)
3
gives 0.2 mole of Al
3+
and 0.3 mole of SO
4
2−
.
∴ Total no of moles of ions present in 1 litre of solution =0.1+0.1+0.2+0.3=0.7 M
0.1 M of CuSO
4
gives 0.2 eq.of Cu
2+
and 0.2 eq.of SO
4
2−
.
Also, 0.1 M of Al
2
(SO
4
)
3
gives 0.6 eq. of Al
3+
and 0.6 eq.of SO
4
2−
.
∴ Total Eq.= 0.2+0.2+0.6+0.6=1.6 N
lanation:
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